A load of mas m = 300kg is lifted by a winch with an acceleration `a = 2 m/s^(2)`. Find the work done during the first one and a half seconds from the beginning of motion.

Explanation

The height to which the body is lifted during the first t second is `h = 1/2 at^(2)`
The tension in the rope is given by ` T = mg + ma`
Work done is given by
`W= Th - F_(net).S. -m(g + a) . 1/2 at^(2)`
Here ` m = 300 kg, a = 2 m/s^(2), g = 10 m/s^(2) t = 1.5 s`
`therefore W = (300)(10 + 2)[1/2(2)(1.5)^(2)]` or `W = 8.1 kJ`